Thevenin Equivalent Impedances Using Zbus
Abstract
A technique for finding the Thevenin equivalent between any pair of nodes of a circuit with a single node-voltage
analysis (plus evaluation of a simple formula for pairs where one of the nodes is not the reference node), and a matrix
inversion is well known to most power-system engineers, but not to most electronic engineers. The technique involves
writing the circuit equations as Y $ X = S . Where Y is called the admittance matrix, X is the vector of unknowns
consisting of the node-voltages (in order) followed the voltage-source currents (also in order), and S is the vector of
right-hand sides of the node-voltage equations (often called the source vector). Once the node-voltages have been
calculated, one calculates
Z = Y K1. Then the Thevenin equivalent impedance between node i and the reference node is found as
ZTh = Zi, i and the Thevenin equivalent impedance between node i and node j is found as
i
ZTh = Zi, i CZj, j K Zi, j KZj, i . Of course the Thevenin equivalent voltages are simply the node-voltages calculated
i, j
above. A detailed example is given, wherein the Thevenin equivalent impedances are calculated as above and compared
to the Thevenin equivalent impedances as calculated by the conventional method for each node to ground and for one
representative example in which neither i nor j is the reference node. This is accomplished without the need for supernodes.
The Node Voltage Method Without Super-nodes
The connection of two or more components is called a node. A component connected between two nodes is called a
branch.
1. Identify (but do not label) all the nodes of the circuit with a visible dot.
2. Choose a reference node (In electronic circuits, this is usually the node on the bottom of the circuit. In power circuits
this is the neutral or the earth-ground.), and label it Node 0.
3. Label each node with a node voltage (even if it is connected to a voltage source) as V1, V2, etc. Write all dependent
source dependencies in terms of the node voltages, the component values and the dependent source gains and place them
on the schematic in the appropriate places.
4. Add a current variable for each voltage source. (Use the passive convention.)
5. Write KVL at each non-reference node in terms of the node voltages, the admittances of the components, the
dependent source gains, and the current variables assigned to the voltage sources in step 4.
6. Write the node-voltage constraints resulting from the voltage sources in current source order.
7. Write the resulting linear equations in matrix form as [Y] [X]=[S], where [Y] is the admittance matrix, [X] is the
vector of node voltages and voltage source currents, and [S] is the source vector (the vector of the right-hand sides of the
equations written in step 5. and 6.). (Some of the entries in [S] will be independent voltage source voltage values, some
of the entries will be independent current source current values, and some of the entries will be zero.)
8. Use a linear solver (a calculator, Maple, Mathematica, Maxima, Matlab, Octave, etc.) to solve the equations from step
7. for the unknown values in [X].
The size of the matrix and the vectors can often be reduced by not assigning node-voltages to the un-grounded terminals
of grounded voltage sources. When the value of a grounded voltage source is needed in a KVL equation its value is used
instead of the node-voltage variable that we have omitted. Treating grounded voltage sources in this manner reduces the
number of variables ( and the number of equations) by two for each voltage source. The variables thus eliminated are the
node-voltage and the current variable associated with the voltage source. The equations removed are the node-voltage
constraint, and the eliminated node-voltage equation.
The size of the matrix and vectors can also be reduced my eliminating any degenerate nodes. A degenerate node is one
that has exactly two impedances - and nothing else - connected to it. For example if one has a resistor of value R
connected from node k to node m and an impedance jX connected from node m to node n, we could eliminate the
degenerate node m by defining only node k and node n with impedance R + jX connected from node k to node n. Of
course, there cannot be anything else connected to node m for this to work.
Example Circuit
Figure 1. Example Circuit
Following steps 1 - 4 above, we have the circuit of Figure 2 below. Note that we have added the node-voltage numbers,
and a current for each voltage source. Note also, that we have slightly simplified the drawing by using the labeled node
facility for node 5. In addition we have written the dependent source depedencies in terms of the node voltages and the
circuit immittances, so it becomes apparent that V3 = 2$V5 , and the current leaving node 5 through the current source is
V2 KV3
3$
. This is called "Doing the algebra on the circuit diagram," by Dr. David Middlebrook at Cal Tech.
8k
Figure 2. Example Circuit with Node-voltages and Source-dependencies in Terms of Node-voltages
The Node Voltage equations are:
V1 KV2
I1 C
=0
5000
I1 C
V2 KV1
5000
V3 KV2
8000
KI2 C
V2 KV3
C
8000
1
1
13
V2 K
V1 K
V CI = 0
5000
8000 3 2
40000
(2)
1
1
V3 K
V CI = 0
8000
8000 2 3
(3)
CI3 = 0
V4 KV5
6000
V2 KV3
8000
(1)
CI2 = 0
=0
KI2 C
3$
1
1
V1 K
V =0
5000
5000 2
C
V5 KV4
6000
C
1
1
V4 K
V =0
6000
6000 5
(4)
V5
=0
10000
3
3
1
1
V2 K
V3 C
V5 K
V =0
8000
8000
3750
6000 4
(5)
The constraint equation associated with I1 is
V1 = 10
V1 = 10
(6)
V2 KV4 = 20
(7)
V3 = 2 V5
(8)
The constraint equation associated with I2 is
V2 KV4 = 20
The constraint equation associated with I3 is
V3 = 2$V5
Since Maple has collected the equations on the analysis variables(the node-voltages and the voltage-source currents), it is
easy to write [Y], [X], and [S]. (Of course, is simple enough to write [Y], [X], and [S] directly from the un-collected
equations as we have done. This is most typically done when using a calculator to solve the matrix equation.)
YBus is then, written as
1
5000
1
5000
K
0
Yd
1
5000
0
0
0
1
0
0
0
0
0
1
0
1
8000
0
0
0
0
1
0
1
6000
K
1
1
C
5000
8000
1
8000
K
0
0
0
3
8000
1
0
0
0
1
8000
K
1
6000
K
0 K1 0
1
K
6000
1
1
C
10000
6000
0
0
0
0
0
0
0
0
0
1
0
K1
0
0
0
0
0
1
0
K2
0
0
0
3
8000
K
1
5000
1
5000
K
0
V1
V2
V3
X=
V4
V5
I1
I2
I3
0
0
0
Sd
0
0
10
20
0
1
5000
K
13
40000
1
8000
K
0
0
0
1
0
0
1
K
8000
0
0
0
1
0
1
8000
0
0
0
0
1
1
6000
0
0
0
0
3
8000
3
K
8000
1
0
0
0
1
0
0
1
6000
1
6000
K
0 K1 0
1
3750
0
0
0
0
0
0
0
0
0
K1
0
0
0
0
1
0
K2
0
0
0
K
(9)
0
0
0
0
(10)
0
10
20
0
NV d evalf LinearAlgebra LinearSolve Y, S
10.
26.29834254
36.46408840
6.298342541
(11)
18.23204420
0.003259668508
K0.001988950276
K0.001270718232
K1
ZBus = YBus
Z d YK1
0
0
0
0
0
580000
181
0
580000
181
K
0
500000
181
0
500000
181
K
0
580000
181
0
580000
181
K
0
250000
181
0
250000
181
K
1
116
181
0
116
181
0
55
181
0
0
10
181
1
126
181
K
10
181
0
1
0
500000
181
116
181
180
181
K
1180000
181
100
181
280
181
K
500000
181
116
181
590000
181
50
181
100
181
K
13
181000
K
1
181
K
140
181
9
45250
15
181
11
181000
K
85
181
1
90500
K
0
115
181
199
181
115
181
K
190
181
K
23
181000
K
47
362000
1
14480
1
14480
21
362000
(12)
We now show that the Thevenin equivalent resistances calculated above are correct, by calculating each equivalent
resistance between a node and the reference node from first principles. We also present one example of calculating the
Thevenin equivalent resistance between one pair of non-reference nodes.
The Thevenin Equivalent Resistance Between Node 1 and the Reference Node
We replace the independent voltage sources with short circuits, and excite the circuit with a one Ampere current source
at node one. The resulting voltage at node 1 is the Thevenin equivalent resistance we seek.
Figure 3. Circuit for Calculation of the Thevenin Equivalent Resistance Between Node 1 and Node 0
We can see from the diagram that the impedance between node 1 and the reference node is zero.
Note that we have followed the advice given above and have not written KVL at the un-grounded node of a grounded
voltage source, reducing the number of node-voltage equations to two. The node voltage equations are then:
VA
5000
C
VB KVA
6000
VA K2$ VB
8000
C
C
VB
10000
C3$
VA KVB
6000
=0
VA K2$ VB
8000
or multiplying both sides by 1000 gives
VA
5
C
VA K2$ VB
VB KVA
6
8
C
VB
10
C
C3$
VA KVB
6
=0
VA K2$ VB
8
=0
=0
Y1 d
1
1
1
C C
5
8
6
1
3
C
6
8
K
S1 d
1
1
K
4
6
K
1
1
3
C
K
6
10 4
59
120
K
5
12
5
24
29
K
60
(13)
0
0
0
0
(14)
NV1 d evalf LinearAlgebra LinearSolve Y1, S1
0.
0.
ZTh
1, 0
(15)
= VA = NV11 = 0.
Comparing to the value of Z1, 1 we have
evalf Z1, 1 = 0.
The Thevenin equivalent between node 1 and the reference node is V1 in series with 0 Ω , i.e.,
10 V in series with 0 Ω . This is obvious, of course, since node 1 has an ideal voltage source connected to it. We include
this case in the example to verify the method.
The Thevenin Equivalent Resistance Between Node 2 and the Reference Node
Figure 3. Circuit for Calculation of the Thevenin Equivalent Resistance Between Node 2 and Node 0
The node voltage equations are:
VA
5000
C
VB KVB
6000
VA
5
C
C
8000
VB
C
10000
VA K2$VB
VB KVB
6
VA K2$VB
8
VB
C
10
VA KVB
6000
VA KVB
C3$
C
C3$
=1
8000
VA KVB
6
or
= 1000
VA KVB
8
=0
=0
which gives
Y2 d
1
1
1
C C
8
6
5
1
3
C
6
8
K
1
1
K
6
4
K
1
1
3
C
K
6
10
4
59
120
K
5
12
5
24
29
K
60
(16)
1000
S2 d
0
1000
0
(17)
NV2 d evalf LinearAlgebra LinearSolve Y2, S2
3204.419890
1381.215470
ZTh
2, 0
= VA = NV21 = VA = 3204.419890
evalf Z2, 2 = 3204.419890
So the Thevenin equivalent between node 2 and the reference node is V2 = 26.29834254 in series with
3204.419890 Ω
The Thevenin Equivalent Resistance Between Node 3 and the Reference Node
Figure 4. Circuit for Calculation of the Thevenin Equivalent Resistance Between Node 3 and Node 0
VA
5000
C
VB KVA
6000
VA KVB
6000
C
C
VB
10000
VA K2$ VB
C3$
8000
=0
VA K2$ VB
8000
=0
(18)
1
1
1
C C
5
6
8
Y3 d
1
3
C
6
8
K
1
1
K
6
4
K
1
1
3
C
K
6
10
4
59
120
K
5
12
5
24
29
K
60
0
S3 d
0
0
(20)
0
NV3 d evalf LinearAlgebra LinearSolve Y3, S3
ZTh
3, 0
(19)
=
0.
0.
= 2$ VB = 0
Z3, 3 = 0
So the Thevenin equivalent between node 3 and the reference node is V3 = 36.46408840 in series with
0Ω.
The Thevenin Equivalent Resistance Between Node 4 and the Reference Node
Figure 5. Circuit for Calculation of the Thevenin Equivalent Resistance Between Node 4 and Node 0
This is the same circuit as for ZTh
2
VA
5
C
VA K2$VB
8
VB KVA
6
Y4 d
VB
C
10
C
VA KVB
C3$
1
3
C
6
8
= 1000
VA K2$ VB
8
1
1
1
C C
5
8
6
K
S4 d
6
=0
1
1
K
4
6
K
1
1
3
C
CK
6
10
4
59
120
K
5
12
5
24
29
K
60
(21)
1000
0
1000
0
(22)
NV4 d evalf LinearAlgebra LinearSolve Y4, S4
3204.419890
1381.215470
ZTh = VA = 3204.419890
4
evalf Z4, 4 = 3204.419890
So the Thevenin equivalent between node 4 and the reference node is V4 = 6.298342541 in series with
3204.419890 Ω .
The Thevenin Equivalent Resistance Between Node 5 and the Reference Node
(23)
Figure 6. Circuit for Calculation of the Thevenin Equivalent Resistance Between Node 5 and Node 0
VA
5
C
VA K2$VB
8
VB KVA
6
Y5 d
VB
C
10
C
C3$
VA KVB
6
VA K2$ VB
1
1
1
C C
5
8
6
1
3
C
6
8
K
S5 d
=0
8
= 1000
1
1
K
4
6
K
1
1
3
C
K
6
10
4
59
120
K
5
12
5
24
29
K
60
(24)
0
1000
0
1000
(25)
NV5 d evalf LinearAlgebra LinearSolve Y5, S5
K2762.430939
K3259.668508
ZTh = VB =K3259.668508
5
(26)
evalf Z5, 5 = K3259.668508
So the Thevenin equivalent between node 5 and the reference node is V5 = 18.23204420 in series with
K3259.668508 Ω .
The Thevenin Equivalent Resistance Between Two Un-grounded Nodes
ZTh = Zi, i CZj, j K Zi, j KZj, i
i, j
Note that the theorem is often quoted as
ZTh = Zi, i CZj, j K2$Zi, j
i, j
But this is true if and only if YBus and ZBus are symmetric. (Most power systems have symmetric matrices, but most
electronic circuits do not.)
The Thevenin Equivalent Resistance Between Node 2 and Node 5
Figure 7. Circuit for Calculation of the Thevenin Equivalent Resistance Between Node 2 and Node 5
The node voltage equations are:
VA
5
C
VA K2$VB
8
C
VA KVB
6
= 1000
VB
10
C
Y25 d
VB KVA
6
VA K2$VB
1
1
1
C C
5
8
6
1
3
C
6
8
K
S25 d
$3$
8
=K1000
1
1
K
4
6
K
1
1
3
C K
10
6
4
59
120
K
5
12
5
24
29
K
60
(27)
1000
K1000
1000
(28)
K1000
NV25 d evalf LinearAlgebra LinearSolve Y25, S25
5966.850829
(29)
4640.883978
NV251 KNV252 = 1325.966851
So the Thevenin equivalent between node 2 and node 5 is evalf NV2 KNV5
1325.966851 Ω .
ZTh
2, 5
= 8.066298343 in series with
= evalf Z2, 2 CZ5, 5 K Z2, 5 KZ5, 2 = 1325.966851
We have presented a technique for finding the Thevenin equivalent between any pair of nodes of a circuit with a single
node-voltage analysis (plus evaluation of a simple formula for pairs where one of the nodes is not the reference node),
and a matrix inversion. The presence of a matrix inversion should pose no computational difficulties for even very large
systems. Maple(14) was able to invert a random, 900 by 900 matrix of IEEE 64-bit floating point numbers in less than
10 seconds on a 2.4 GHz, p8600 laptop with 3GB of 789 MHz RAM.